Let $H$ be a separable Hilbert space and let $\operatorname{U}(H)$ be the group of unitary transformations of $H$. It is well known that the weak, strong and compact-open topologies on $\operatorname{U}(H)$ all agree (Espinoza and Uribe - Topological properties of the unitary group).
In particular, they all induce the standard topology on the center $\operatorname{U}(1) \subset \operatorname{U}(H)$, and the quotient $P{\operatorname{U}(H)} = \operatorname{U}(H)/{\operatorname{U}(1)}$ is a topological group.
Now, $P{\operatorname{U}(H)}$ is the automorphism group of the $C^*$-algebra $A = \operatorname{B}(H)$, and for generic $C^*$-algebras, one usually considers the "point-norm-topology" on the automorphism group $\operatorname{Aut}(A)$, which is just the restriction of the strong topology on space of all bounded operators on $A$.
On the other hand, since $\operatorname{B}(H)$ is a von Neumann algebra, the more natural topology may be Haagerup's u-topology (see Haagerup - The standard form of von Neumann algebras). By Corollary 3.8 in Haagerup's paper, for $\operatorname{B}(H)$ this coincides with the p-topology, which is the restriction of the locally convex topology on the set of all $\sigma$-weakly bounded transformations of $A$ given by the collection of seminorms $$\varphi \mapsto \lvert\xi(\varphi(a))\rvert, ~~\xi \in A_* ,a \in A, \varphi \in \operatorname{Aut}(A).$$ Here $A_*$ denotes the predual, which in our case is the space of trace-class operators.
Q: Do these two topologies on $P{\operatorname{U}(H)}$ also coincide, and are they equal to the quotient topology of the weak/strong topology?
